Integrand size = 26, antiderivative size = 411 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\frac {6 b^3 f m n^3 \log (x)}{e}-\frac {6 b^2 f m n^2 \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {3 b f m n \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e}-\frac {f m \log \left (1+\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^3}{e}-\frac {6 b^3 f m n^3 \log (e+f x)}{e}-\frac {6 b^3 n^3 \log \left (d (e+f x)^m\right )}{x}-\frac {6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x}+\frac {6 b^3 f m n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{e}+\frac {6 b^2 f m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{e}+\frac {3 b f m n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{e}+\frac {6 b^3 f m n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{e}+\frac {6 b^2 f m n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{e}+\frac {6 b^3 f m n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x}\right )}{e} \]
6*b^3*f*m*n^3*ln(x)/e-6*b^2*f*m*n^2*ln(1+e/f/x)*(a+b*ln(c*x^n))/e-3*b*f*m* n*ln(1+e/f/x)*(a+b*ln(c*x^n))^2/e-f*m*ln(1+e/f/x)*(a+b*ln(c*x^n))^3/e-6*b^ 3*f*m*n^3*ln(f*x+e)/e-6*b^3*n^3*ln(d*(f*x+e)^m)/x-6*b^2*n^2*(a+b*ln(c*x^n) )*ln(d*(f*x+e)^m)/x-3*b*n*(a+b*ln(c*x^n))^2*ln(d*(f*x+e)^m)/x-(a+b*ln(c*x^ n))^3*ln(d*(f*x+e)^m)/x+6*b^3*f*m*n^3*polylog(2,-e/f/x)/e+6*b^2*f*m*n^2*(a +b*ln(c*x^n))*polylog(2,-e/f/x)/e+3*b*f*m*n*(a+b*ln(c*x^n))^2*polylog(2,-e /f/x)/e+6*b^3*f*m*n^3*polylog(3,-e/f/x)/e+6*b^2*f*m*n^2*(a+b*ln(c*x^n))*po lylog(3,-e/f/x)/e+6*b^3*f*m*n^3*polylog(4,-e/f/x)/e
Leaf count is larger than twice the leaf count of optimal. \(1347\) vs. \(2(411)=822\).
Time = 0.40 (sec) , antiderivative size = 1347, normalized size of antiderivative = 3.28 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx =\text {Too large to display} \]
-1/4*(-4*a^3*f*m*x*Log[x] - 12*a^2*b*f*m*n*x*Log[x] - 24*a*b^2*f*m*n^2*x*L og[x] - 24*b^3*f*m*n^3*x*Log[x] + 6*a^2*b*f*m*n*x*Log[x]^2 + 12*a*b^2*f*m* n^2*x*Log[x]^2 + 12*b^3*f*m*n^3*x*Log[x]^2 - 4*a*b^2*f*m*n^2*x*Log[x]^3 - 4*b^3*f*m*n^3*x*Log[x]^3 + b^3*f*m*n^3*x*Log[x]^4 - 12*a^2*b*f*m*x*Log[x]* Log[c*x^n] - 24*a*b^2*f*m*n*x*Log[x]*Log[c*x^n] - 24*b^3*f*m*n^2*x*Log[x]* Log[c*x^n] + 12*a*b^2*f*m*n*x*Log[x]^2*Log[c*x^n] + 12*b^3*f*m*n^2*x*Log[x ]^2*Log[c*x^n] - 4*b^3*f*m*n^2*x*Log[x]^3*Log[c*x^n] - 12*a*b^2*f*m*x*Log[ x]*Log[c*x^n]^2 - 12*b^3*f*m*n*x*Log[x]*Log[c*x^n]^2 + 6*b^3*f*m*n*x*Log[x ]^2*Log[c*x^n]^2 - 4*b^3*f*m*x*Log[x]*Log[c*x^n]^3 + 4*a^3*f*m*x*Log[e + f *x] + 12*a^2*b*f*m*n*x*Log[e + f*x] + 24*a*b^2*f*m*n^2*x*Log[e + f*x] + 24 *b^3*f*m*n^3*x*Log[e + f*x] - 12*a^2*b*f*m*n*x*Log[x]*Log[e + f*x] - 24*a* b^2*f*m*n^2*x*Log[x]*Log[e + f*x] - 24*b^3*f*m*n^3*x*Log[x]*Log[e + f*x] + 12*a*b^2*f*m*n^2*x*Log[x]^2*Log[e + f*x] + 12*b^3*f*m*n^3*x*Log[x]^2*Log[ e + f*x] - 4*b^3*f*m*n^3*x*Log[x]^3*Log[e + f*x] + 12*a^2*b*f*m*x*Log[c*x^ n]*Log[e + f*x] + 24*a*b^2*f*m*n*x*Log[c*x^n]*Log[e + f*x] + 24*b^3*f*m*n^ 2*x*Log[c*x^n]*Log[e + f*x] - 24*a*b^2*f*m*n*x*Log[x]*Log[c*x^n]*Log[e + f *x] - 24*b^3*f*m*n^2*x*Log[x]*Log[c*x^n]*Log[e + f*x] + 12*b^3*f*m*n^2*x*L og[x]^2*Log[c*x^n]*Log[e + f*x] + 12*a*b^2*f*m*x*Log[c*x^n]^2*Log[e + f*x] + 12*b^3*f*m*n*x*Log[c*x^n]^2*Log[e + f*x] - 12*b^3*f*m*n*x*Log[x]*Log[c* x^n]^2*Log[e + f*x] + 4*b^3*f*m*x*Log[c*x^n]^3*Log[e + f*x] + 4*a^3*e*L...
Time = 0.77 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2825, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 2825 |
| \(\displaystyle -f m \int \left (-\frac {6 b^3 n^3}{x (e+f x)}-\frac {6 b^2 \left (a+b \log \left (c x^n\right )\right ) n^2}{x (e+f x)}-\frac {3 b \left (a+b \log \left (c x^n\right )\right )^2 n}{x (e+f x)}-\frac {\left (a+b \log \left (c x^n\right )\right )^3}{x (e+f x)}\right )dx-\frac {6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x}-\frac {6 b^3 n^3 \log \left (d (e+f x)^m\right )}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x}-f m \left (-\frac {6 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {6 b^2 n^2 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e}+\frac {6 b^2 n^2 \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {3 b n \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e}+\frac {3 b n \log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e}+\frac {\log \left (\frac {e}{f x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{e}-\frac {6 b^3 n^3 \operatorname {PolyLog}\left (2,-\frac {e}{f x}\right )}{e}-\frac {6 b^3 n^3 \operatorname {PolyLog}\left (3,-\frac {e}{f x}\right )}{e}-\frac {6 b^3 n^3 \operatorname {PolyLog}\left (4,-\frac {e}{f x}\right )}{e}+\frac {6 b^3 n^3 \log (e+f x)}{e}-\frac {6 b^3 n^3 \log (x)}{e}\right )-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{x}-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x}-\frac {6 b^3 n^3 \log \left (d (e+f x)^m\right )}{x}\) |
(-6*b^3*n^3*Log[d*(e + f*x)^m])/x - (6*b^2*n^2*(a + b*Log[c*x^n])*Log[d*(e + f*x)^m])/x - (3*b*n*(a + b*Log[c*x^n])^2*Log[d*(e + f*x)^m])/x - ((a + b*Log[c*x^n])^3*Log[d*(e + f*x)^m])/x - f*m*((-6*b^3*n^3*Log[x])/e + (6*b^ 2*n^2*Log[1 + e/(f*x)]*(a + b*Log[c*x^n]))/e + (3*b*n*Log[1 + e/(f*x)]*(a + b*Log[c*x^n])^2)/e + (Log[1 + e/(f*x)]*(a + b*Log[c*x^n])^3)/e + (6*b^3* n^3*Log[e + f*x])/e - (6*b^3*n^3*PolyLog[2, -(e/(f*x))])/e - (6*b^2*n^2*(a + b*Log[c*x^n])*PolyLog[2, -(e/(f*x))])/e - (3*b*n*(a + b*Log[c*x^n])^2*P olyLog[2, -(e/(f*x))])/e - (6*b^3*n^3*PolyLog[3, -(e/(f*x))])/e - (6*b^2*n ^2*(a + b*Log[c*x^n])*PolyLog[3, -(e/(f*x))])/e - (6*b^3*n^3*PolyLog[4, -( e/(f*x))])/e)
3.1.88.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q* (a + b*Log[c*x^n])^p, x]}, Simp[Log[d*(e + f*x^m)^r] u, x] - Simp[f*m*r Int[x^(m - 1)/(e + f*x^m) u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m , n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 151.33 (sec) , antiderivative size = 16532, normalized size of antiderivative = 40.22
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{2}} \,d x } \]
integral((b^3*log(c*x^n)^3 + 3*a*b^2*log(c*x^n)^2 + 3*a^2*b*log(c*x^n) + a ^3)*log((f*x + e)^m*d)/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{2}} \,d x } \]
-((b^3*f*m*x*log(f*x + e) - b^3*f*m*x*log(x) + b^3*e*log(d))*log(x^n)^3 + (b^3*e*log(x^n)^3 + 3*(e*n + e*log(c))*a^2*b + 3*(2*e*n^2 + 2*e*n*log(c) + e*log(c)^2)*a*b^2 + (6*e*n^3 + 6*e*n^2*log(c) + 3*e*n*log(c)^2 + e*log(c) ^3)*b^3 + a^3*e + 3*((e*n + e*log(c))*b^3 + a*b^2*e)*log(x^n)^2 + 3*(2*(e* n + e*log(c))*a*b^2 + (2*e*n^2 + 2*e*n*log(c) + e*log(c)^2)*b^3 + a^2*b*e) *log(x^n))*log((f*x + e)^m))/(e*x) + integrate((b^3*e^2*log(c)^3*log(d) + 3*a*b^2*e^2*log(c)^2*log(d) + 3*a^2*b*e^2*log(c)*log(d) + a^3*e^2*log(d) + 3*(a*b^2*e^2*log(d) + (e^2*n*log(d) + e^2*log(c)*log(d))*b^3 + ((e*f*m + e*f*log(d))*a*b^2 + (e*f*m*n + e*f*n*log(d) + (e*f*m + e*f*log(d))*log(c)) *b^3)*x + (b^3*f^2*m*n*x^2 + b^3*e*f*m*n*x)*log(f*x + e) - (b^3*f^2*m*n*x^ 2 + b^3*e*f*m*n*x)*log(x))*log(x^n)^2 + ((e*f*m + e*f*log(d))*a^3 + 3*(e*f *m*n + (e*f*m + e*f*log(d))*log(c))*a^2*b + 3*(2*e*f*m*n^2 + 2*e*f*m*n*log (c) + (e*f*m + e*f*log(d))*log(c)^2)*a*b^2 + (6*e*f*m*n^3 + 6*e*f*m*n^2*lo g(c) + 3*e*f*m*n*log(c)^2 + (e*f*m + e*f*log(d))*log(c)^3)*b^3)*x + 3*(b^3 *e^2*log(c)^2*log(d) + 2*a*b^2*e^2*log(c)*log(d) + a^2*b*e^2*log(d) + ((e* f*m + e*f*log(d))*a^2*b + 2*(e*f*m*n + (e*f*m + e*f*log(d))*log(c))*a*b^2 + (2*e*f*m*n^2 + 2*e*f*m*n*log(c) + (e*f*m + e*f*log(d))*log(c)^2)*b^3)*x) *log(x^n))/(e*f*x^3 + e^2*x^2), x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d (e+f x)^m\right )}{x^2} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x^2} \,d x \]